We start with a function which we assume to be Riemann-Stieltjes integrable by the function . Now, instead of the full generality we used before, let’s just let be a strictly increasing continuous function with and . Define and to be the composite functions and . Then is Riemann-Stieltjes integrable by on , and we have the equality

For decreasing functions we get almost the exact same thing, so you should figure out the parallel statement and proof yourself.

Since is strictly increasing, it must be one-to-one, and it’s onto by assumption. In fact, is an explicit homeomorphism of the intervals and , and its inverse is also a strictly increasing continuous function. We can now use and its inverse to set up a bijection between partitions of and : if is a partition, then is a partition, and vice versa. Further, refinements of partitions of one side correspond to refinements of partitions on the other side.

So if we’re given an then there’s some partition of so that for any finer partition we have . Let be the corresponding partition of , and let be a partition of finer than it. Then it’s easily verified that the Riemann-Stieltjes sum is equal to the Riemann-Stieltjes sum . Everything else follows quickly from here.

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I can’t resist making a few comments on Riemann-Stieltjes from the point of view of functional analysis:

(1) The Riemann-Stieltjes integral is defined whenever is continuous on and is of bounded variation on .

(2) The Riemann-Stieltjes integral is defined for all continuous on only if $\latex \alpha$ is of bounded variation on , and is defined for all of bounded variation on only if is continuous on .

(3) The Riemann-Stieltjes integral satisfies the inequality where is the sup norm and is the total variation. It is therefore a continuous or bounded linear map in each of its separate variables and .

(4) The pairing is “perfect” in the sense that it expresses the Banach space of functions of bounded variation (modulo constants) as the strong dual of the Banach space of continuous functions on . (However, it doesn’t work the other way: the dual of constants is not .)

Normally the space dual to is expressed in terms of measures, so this gives a nice “concrete” way of thinking about measures (for example, Dirac measures are of the form where is a Heaviside function, which is locally constant but for a single jump discontinuity — the size of the jump gives the weight of the measure).

It also sheds some light on things like the Radon-Nikodym theorem: a measure on which is absolutely continuous with respect to Lebesgue measure is of the form where is an absolutely continuous function, and the derivative of (as in ) is an function with respect to Lebesgue measure . Indeed, differentiation gives a Banach space isomorphism .

I find this approach much gentler and more intuitive than that taken in, e.g., Rudin’s Real and Complex Analysis (which is undeniably efficient and frequently very clever, but brutally abstract in the discussion of things like Radon-Nikodym).

Anyhow, these are good points, most of which I’m going to get to eventually. In fact, after I rework integration by parts today I’m going to run through the basics of bounded variation tomorrow through Friday before coming back to more about Riemann-Stieltjes integration once I have that language down.

Oh, and the Radon-Nikodym observation is key. That’s one of the reasons I think that the Riemann-Stieltjes theory (and later the Lebesgue-Stieltjes theory) are really useful for understanding what’s really going on.

[…] unknown wrote an interesting post today onHere’s a quick excerptLast Friday we explained the change of variables formula for Riemann integrals by using Riemann-Stieltjes integrals. Today let’s push it a little further and prove a change of variables formula for Riemann-Stieltjes integrals. … […]

D.W., it’s covered in lots of places (e.g., Measure and Integral by Wheeden and Zygmund). It has likely been covered here somewhere on this blog, too. To prove it, just note that any Riemann-Stieltjes sum approximating the Riemann-Stieltjes integral,

has its norm bounded above by

where is the supremum of over the given interval . By the triangle inequality, this is bounded above by

and this in turn is bounded above by , by definition of the variation .

Thanks a lot. Let me please ask another question.
I have the difference of two integrals over the whole real axis
(int K dFn – int K dF).
Fn and F are both nonnegative increasing functions, K is of bounded variation. Fn is discontinous, F is continous. For K, the limes for |x| to infty is 0.
I would like to make integration by parts, so that the term becomes
int (Fn-F) dK.
Under which assumptions does this work? I think that the integral is not well defined if K is discontinous. But can I be sure that it works if K is continous?

If I understand the question, one will have to pay attention to growth rates as well. For example, if F_n – F grows very rapidly but K(x) tends to 0 slowly as |x| -> \infty, then there is no reason to suppose the integral over the real line will converge. Other than that, I think the main obstruction to the existence of the Riemann-Stieltjes integral will be when the functions F_n – F and K have a common point of discontinuity. In particular, if K is continuous, then under your assumptions we know that F_n – F is of bounded variation on any finite interval [a, b], and therefore the Riemann-Stieltjes integral over that integral exists, and one can perform an integration by parts over that interval without worry. Under further assumptions about growth rates at infinity (e.g. if the product of (F_n – F)(x) and K(x) tends to 0 as |x| -> \infty), the integral over (-\infty, \infty) also exists and again one can perform the integration by parts.

It might be wise to check with your local analyst. I’d have to think a little further to make sure I’m being 100% accurate here, but I think I’m okay.

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